Not signed in (Sign In)

Start a new discussion

Not signed in

Want to take part in these discussions? Sign in if you have an account, or apply for one below

• Sign in using OpenID

Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

• CommentRowNumber1.
• CommentAuthorTobyBartels
• CommentTimeNov 12th 2010

I’ve decided that these shouldn’t exist (making me agree with the standard terminology) and explained why at regular cardinal.

• CommentRowNumber2.
• CommentAuthorMike Shulman
• CommentTimeNov 13th 2010

What is a “nullary copower”?

• CommentRowNumber3.
• CommentAuthorTobyBartels
• CommentTimeNov 13th 2010
• (edited Nov 13th 2010)

I’m not sure, actually.

I want to say that taking indexed disjoint unions is a kind of binary operation; a special case of this is taking a binary cartesian product, where the family (whose disjoint union we’re taking) is constant. So $Set_{\lt\kappa}$ is required to be closed under binary cartesian products and therefore morally also ought to be required to be closed under the nullary cartesian product. So originally I wrote that $Set_{\lt\kappa}$ should have a terminal object.

But then I realised that it is just a coincidence (for some interpretation of ‘just’) that a constant indexed coproduct is a binary product in $Set$. What really matters is not that it’s a product but that it’s a copower. So I wrote ‘nullary copower’. But you’re right; I’m not sure what that actually means. I was borrowing a friend’s computer to record a passing thought and didn’t have time to think more about it. (I don’t now either; my mom has just walked in, and we have to leave for a show.)

For the record, that clause is needed to prevent $1$ from being a regular cardinal; the other clauses prevent $0$ and $2$.

• CommentRowNumber4.
• CommentAuthorSridharRamesh
• CommentTimeNov 13th 2010
• (edited May 8th 2011)

Outside of any context, I would almost always interpret “nullary copower” to mean “initial object” (just as I would interpret “nullary power” to mean “terminal object”), but that of course isn’t actually the sense of “nullary” that you mean here.

I do think I understand the sense of “nullary” that you are getting at. Let me try to give my own account of it. A general, “multi-level” indexed collection of collections of collections of… can be thought of as indexed by a rooted tree, each of whose nodes is labelled either “index” or “item”, with “item” nodes having no children (basically, thinking of the children of any “index” node as the collection indexed by that node). Taking the ultimate disjoint union of this multi-level beast amounts to taking its collection of “item” nodes. If the item nodes of the tree are precisely the nodes at some constant depth h from the root, this is the process of taking the ultimate sum of an indexed collection of indexed collections of indexed collections of [h many levels down]… things. In particular, when h = 2, this is the standard notion of sending a “singly-“indexed collection of collections to its disjoint union [the children of the root are the indices, and the children of each of those comprise the items in the correspondingly indexed collection]. Of course, h needn’t equal 2. For one particularly trivial tree, h is instead 0. And the process of producing the item nodes of this tree (of which there happens to be just 1, although this is the “just a coincidence”) is the 0-level disjoint union; aka, what you called (but I wouldn’t) the “nullary copower”.

Maybe that doesn’t make anything clearer to anyone except me. But it’s how I would think about it.

• CommentRowNumber5.
• CommentAuthorSridharRamesh
• CommentTimeNov 13th 2010
• (edited Nov 13th 2010)

So, along these lines, I would define k to be regular if every tree of this sort at which every index has < k children has ultimately < k item nodes. This would automatically rule out the regularity of 0, 1, and 2, leaving only the infinite regular cardinals.

Edit: Whoops, it doesn’t rule out 2! Well, alright…

• CommentRowNumber6.
• CommentAuthorSridharRamesh
• CommentTimeNov 13th 2010
• (edited Nov 13th 2010)

Well, I think I’d personally be ok with not ruling that out. It seems to me right now (though I may be making some dumb mistakes; I haven’t thought this all the way through) that my intuition for regularity is something like that it is interesting to study full subcategories N of Set with the property that A) N is closed under finite limits [the structure needed to interpret sets of sets of sets of …, using slice categories], and B) the slice category over any object X in N is indeed equivalent to the category N^X. We might call such an N “regular”, and then describe a cardinal k as regular just in case the full subcategory of sets of cardinality < k is regular.

On this account, the full subcategory of truth values (in the sense of sets whose elements are all equal) is indeed regular, so we should expect that 2 is regular (since being < 2 amounts, at least classically, to being <= 1).

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeNov 13th 2010

I don’t understand why you would want to rule out 0, 1, and 2. If the most natural definition rules them out, fine, if it includes them, fine too. I used to think the most natural definition was closure under indexed coproducts/unions, which includes them. Since binary products are, as you say, “coincidentally” a special case of indexed coproducts, that sort of closure includes closure under binary products, but since terminal objects are not such a special case, it doesn’t include them.

Sridhar, I don’t understand why your definition rules out 1 either. Surely a tree in which every index node has zero children cannot have any item nodes? Unless you are allowing a single item node that is not the child of any index node to count as a “tree,” which seems unnatural to me: I would expect the root should always be an index node.

Re: “classically,” I think that since constructively (or even without AC), cardinal numbers (= isomorphism classes of sets) are not even totally ordered, instead of talking about regular cardinals one should talk about regular families of cardinalities, at least in cases where what one is really interested in classically is the closure properties of the cardinalities $\lt\kappa$.

• CommentRowNumber8.
• CommentAuthorSridharRamesh
• CommentTimeNov 13th 2010
• (edited Nov 13th 2010)

I agree, one should really talk about regular families of cardinalities. I’m all for it. I gave above my proposed definition of the conditions under which a full (and, if you like, replete) subcategory of Set should count as regular, which seemed quite nice, though I fear I will discover a fly in the ointment.

Concerning the presentation in terms of trees, I had been allowing the root node to be an item node. My notion of tree was essentially the type inductively defined by “A tree is either the particular tree ’Item’ or an indexed collection of trees”. Perhaps this should be modified, but it seems to me a very natural recursive definition (with a correspondingly natural recursive definition of the “disjoint sum” of a tree/the collection of items in a tree).

If we force to root of a tree to be an index node, this amounts to instead saying “A tree is an indexed collection of (either ’Item’ or a tree)s”. I suppose that’s not so terribly different, being just as much a natural recursive definition, just with the two constructors in the other order, yet somehow, it strikes me as less right the notion to use here.

In other words, if we think of F(X) as the type of indexed collections of Xes [with the indices restricted to coming from whatever family of cardinalities one likes], I was thinking of the initial algebra of 1 + F(-), whereas the modified notion (where root nodes have to be index nodes) would amount to the initial algebra of F(1 + -). For whatever reason, the former seems [or seemed; my thoughts are now in flux] much more “right” to me, though I don’t seem articulate enough right now to explain why.

• CommentRowNumber9.
• CommentAuthorTobyBartels
• CommentTimeNov 15th 2010

Certainly we really want regular families of cardinals. It’s just that the traditional term for such a thing is ‘regular cardinal’, because it’s also traditional to (use Choice to) think of such families in terms of the smallest cardinal not in the family.

I think that Sridhar’s argument to allow the root to be an item node, so both $0$ and $1$ would be ruled out by the ‘nullary copower’ rule. I’d still like to find the proper term for ‘nullary copower’; we are using the structure of $Set_\kappa$ (or $N$) as indexed over itself, and I don’t really know much about indexed categories. However, if we cheat a bit and instead use the structure of $N$ as a closed category (even though in most cases it is not), then what I called the ‘nullary copower’ is simply the unit object of the closed category. (So the coincidence that makes this a terminal object is that $N$ is cartesian closed, or ‘cartesian self-indexed’ in some sense if we don’t cheat.)

I don’t want to rule out the finite regular cardinals just because I want regular cardinals to be infinite; I agree that we should see where a natural definition leads us. I’m just thinking that if one wants indexed disjunction, then one probably also wants finitary disjunction. It obviously could go either way, and the examples of infinite regular cardinals won’t distinguish these.

• CommentRowNumber10.
• CommentAuthorTobyBartels
• CommentTimeNov 15th 2010

@ Sridhar #6

To what extent do we have a theorem that your conditions (A,B) really are equivalent to the requirement that $N$ be $Set_{\lt\kappa}$ for some regular cardinal (in the classical sense) $\kappa$? In particular, assuming Choice and ruling out finite cardinals by fiat, are they indeed equivalent?

I want to see where the ‘closure under smaller cardinals’ comes in, particularly whether it is closure under subobjects or quotient objects. (I used quotient objects in the article, since constructively the collection of Kuratowski-finite cardinals ought to be regular; there are many conditions made for these cardinals that are not made for subfinite cardinals, particularly conditions of closure under ‘finitary’ operations.)

• CommentRowNumber11.
• CommentAuthorSridharRamesh
• CommentTimeNov 16th 2010

Quoth Toby:

I think that Sridhar’s argument to allow the root to be an item node, so both 0 and 1 would be ruled out by the ‘nullary copower’ rule.

Sorry, I’m having difficulty parsing this. Did part of the sentence go missing?

• CommentRowNumber12.
• CommentAuthorMike Shulman
• CommentTimeNov 17th 2010

I think I don’t want to spend any more time on the definition of “regular cardinal”. What I gather from the discussion so far is that there are different ways of phrasing it, some of which would include some of 0,1,2 and some of which wouldn’t. So I like the current phrasing of the page, which says essentially that.

However I still don’t understand what a nullary copower is supposed to mean, even intuitively. In my mind, a copower is not something that has an arity, so I don’t see how it could be nullary. Unless you mean k-fold iterated copowers, in which case the nullary copower of any object would just be itself.

• CommentRowNumber13.
• CommentAuthorTobyBartels
• CommentTimeNov 20th 2010

@ Sridhar

Did part of the sentence go missing?

Yes, it should have read

I agree with Sridhar’s argument …

or something like that.

@ Mike

Unless you mean k-fold iterated copowers, in which case the nullary copower of any object would just be itself.

Yes, I think that this is what I’m trying to get at, but in a way in which ordinary copowers are binary and its the unary copower of any object that is itself.

The copowers at copower seem to me to be the wrong generality, and I want to look at something in CWM tomorrow. Then I think that I can write down a definition of what I mean, and we can see what it’s really called.

• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeDec 19th 2010

In thinking about familial regularity and exactness, I realized that what I really want out of a “regular family of cardinalities” in that context is that

• it contains 1, and
• if it contains λ, then it contains a sum $\sum_{i\in\lambda} \mu_i$ if and only if it contains each $\mu_i$.

Classically, this implies it is either the set of cardinal numbers less than some infinite regular cardinal, or 2={0,1}, or the set {1} which is not even down-closed. So at least for that application, I agree that 2 should be a regular cardinal, but 0 and 1 should not be—but I also want to include the non-cardinal set {1}.

• CommentRowNumber15.
• CommentAuthorTobyBartels
• CommentTimeDec 20th 2010

I’ve thought about $\{1\}$ as well.

So mine is a strong list of conditions for a regular cardinal (thought of as a set of cardinals which a priori might not itself be a cardinal, even classically):

• closed under quotients,
• closed under finitary sums,
• closed under $1$ and indexed sums (with the index also from the collection).

While yours is a weak list of conditions which happen to imply my list (at least classically) if the regular cardinal is infinite.

Your list has the good point that it’s actually motivated by a specific example.

• CommentRowNumber16.
• CommentAuthorMike Shulman
• CommentTimeDec 21st 2010

I added a paragraph about my list to regular cardinal.

• CommentRowNumber17.
• CommentAuthorMike Shulman
• CommentTimeDec 21st 2010

I’m still wondering what a “nullary copower” is, by the way.

• CommentRowNumber18.
• CommentAuthorTobyBartels
• CommentTimeDec 22nd 2010
• (edited Dec 22nd 2010)

@ Mike #16:

Your paragraph is less abstract, so I moved it higher up. I also put in a bit about $\{1\}$ (in the weak foundations bit, since you only recognise it as a possibility when you’re thinking along those lines, at least at first).

@ Mike #17:

Ah, yes, I should figure that out (and then we can figure out what to call it).

• CommentRowNumber19.
• CommentAuthorMike Shulman
• CommentTimeDec 22nd 2010

Should “if we require only clauses (1&2), then 2 is a regular cardinal” be clauses (1&3)?

• CommentRowNumber20.
• CommentAuthorTobyBartels
• CommentTimeDec 22nd 2010

It should be (1&4). I’ve fixed it.

PS: Please see this new thread.

• CommentRowNumber21.
• CommentAuthorTobyBartels
• CommentTimeJan 29th 2011

In Categories Work (2nd ed, p 64), Mac Lane defines the copower $X \cdot b$ to be the coproduct $\coprod_{i\colon X} b$ of a set-indexed constant family of objects; this is a core-functorial (at least) operation $Set \times C \to C$. So if $C$ is $Set$, then this is a binary operation $Set \times Set \to Set$, which happens to be the cartesian product. This happens to have an identity (although really a left identity would be enough), which is the singleton; so this is the “nullary copower” in $Set$.

The stuff at copower is a generalisation of Mac Lane’s copowers to the enriched case. In general, $V$ is a symmetric monoidal closed category, $C$ is enriched over $V$, and the copower is an operation $V \times C \to C$. Generalising my notion of nullary copower, we are left with the unit object in $V$. Of course, it only makes sense to say that a subcategory of $C$ is closed under this operation when $C$ is $V$, as it is here.

So instead of blathering about nullary copowers, I can just ask that $Set_{\kappa}$ have the unit object of the closed structure on $Set$. But how do I explain this as a variation on the theme of closure under coproducts? I see that copowers are a kind of weighted colimit, so is the singleton a kind of weighted colimit too? I am getting lost on that.

• CommentRowNumber22.
• CommentAuthorMike Shulman
• CommentTimeJan 31st 2011

I would say that the singleton is an iterated coproduct of the form $\coprod_{i_1\in I} \coprod_{i_2 \in I_{i_1}} \dots \coprod_{i_n \in I_{i_1,\dots,i_{n-1}}} X_{i_1,\dots,i_n}$ where $n=0$. I think that’s what Sridhar was getting at with the trees above.

• CommentRowNumber23.
• CommentAuthorTobyBartels
• CommentTimeJan 31st 2011

Yes, Sridhar was definitely getting at the right thing, which is as you said. But I still don’t know what to call it … or didn’t, until you said “iterated coproduct”. Now I think that it works! (Although at the last minute I decided to change “coproduct” to “disjoint union”, because it’s something that really only works in $Set$, at least not without going to a higher level of abstraction than just recognising disjoint unions as coproducts.)

Add your comments
• Please log in or leave your comment as a "guest post". If commenting as a "guest", please include your name in the message as a courtesy. Note: only certain categories allow guest posts.
• To produce a hyperlink to an nLab entry, simply put double square brackets around its name, e.g. [[category]]. To use (La)TeX mathematics in your post, make sure Markdown+Itex is selected below and put your mathematics between dollar signs as usual. Only a subset of the usual TeX math commands are accepted: see here for a list.

• (Help)